Singlet and Triplet states: Why is the $S=0$ state defined as it is?

Question

I'm working on some exercise regarding the spin coupling of two electrons. There we have the wavefunctions corresponding to the S values as

$$\begin{align} S = 1: &\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm] S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow) \end{align}$$

I understand that the three wavefunctions for $S=1$ are symmetric and the one for $S=0$ is antisymmetric. My question is, why is the combination with the minus sign is the one for $S=0$?

My thinking is that in a $\uparrow \downarrow$ or $\downarrow\uparrow$ combination the $S_z$ components would already add up to $0$ so that the minus sign would not change anything in the fact that $S=0$.

Or is it that $\downarrow\uparrow$ or $\uparrow\downarrow$ each represent a state with $S=1$ and $S_z = 0$ so that subtracting the one from the other gives $S=1-1=0$?

Answer 1

You are correct that one of the $S = 1$ states has zero angular momentum along the $z$ axis. However, the $S = 0$ state has zero angular momentum along any direction, and this follows directly from the presence of the minus sign.

To see why, consider the operator for spin along a general direction, $S_{\hat{n}}$. Since the spins are identical, it doesn't matter which spin is what, so $$\langle \uparrow \downarrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \alpha(\hat{n}).$$ Moreover, the operator $P$ that interchanges the spins doesn't affect $S_{\hat{n}}$, because $$P S_{\hat{n}} = P \left(S_{\hat{n}}^1 + S_{\hat{n}}^2 \right) = S_{\hat{n}}^2 + S_{\hat{n}}^1 = S_{\hat{n}}$$ which further implies that $$\langle \uparrow \downarrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \alpha(\hat{n}).$$ This is all the information we need to evaluate the spin along the $\hat{n}$ direction for the two states you give. For the $S = 1$ state, we find $$\frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} = 2\alpha$$ while for the $S = 0$ state we find $$\frac{\alpha}{2} + \frac{\alpha}{2} - \frac{\alpha}{2} - \frac{\alpha}{2} = 0.$$ The cross terms pick up a minus sign, due to the minus sign in the superposition. This shows the singlet has zero spin along any direction.

Answer 2

It's important to remember that when we say $S=1, 0$ we're really thinking of the eigenstates and eigenvalues of $\mathbf{S}^2 = (\mathbf{S}_1 + \mathbf{S}_2)^2$. Try applying this operator, and you'll see that the symmetric states have eigenvalue $S=1$, and the antisymmetric will have eigenvalue $S=0$. If you're simply applying $\mathbf{S}_z$, which is what you are doing in your question statement, then definitely both the mixed symmetric and antisymmetric states have eigenvalue $j=0$, as you've pointed out.

Answer 3

Just a slightly different view using the interchange operator $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $

My question is, why is the combination with the minus sign is S=0?

Another way to look at this is:

$$\begin{align} S = 1: &\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm] S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow) \end{align}$$

The triplet states are not affected by the interchange of $ m_1\iff m_2$ whereas the $S=0$ changes sign under interchange.

Using $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $ as the interchange operator.

This results in

$$\hat P \textrm{(triplet)} = \textrm{triplet} $$ But $$\hat P\textrm{(singlet)} = -~\textrm{ singlet} $$